mm-739
===
Subject: Re: Internal Set Theory Uniqueness Principle
> NelsonÕs theorem and proof can be extended to the \
following
result.
> Let P(u,v_1,....,v_k) be a formula in Internal Set Theory
such that
> the only occurrences of the predicate \
ŌstandardÕ in the
formula are
> in quanti\[CapitalThorn]ers of the type for all standard z in V or for
some
> standard z in V for some standard set V (V can differ from
quanti\[CapitalThorn]er
> to quanti\[CapitalThorn]er). If y_1, ..., y_k are standard, and there is
a unique
> x such that P(x,y_1,...,y_k) holds, the unique value of x
is standard.
> My question relates to what happens if for all standard z
in V and
> for some standard z in V are relaxed to for all standard z
and
> for some standard z. Is it still true that x must be
standard?
IÕm thinking this is true because it seems to me every
external formula can
be reduced to a weakly equivalent internal formula, with no
proviso needed.
Indeed, just as in logic, we can arrange to put the external
formula into
an
equivalent prenex form, whose prenex is a sequence of
internal and
external quanti\[CapitalThorn]ers. Using the triviality and its dual that
for all x
commutes with for all standard x and then also using
idealization to
push
the external generalizations to the left of the internal
instantiations and
the external instantiations to the left of the internal
generaliztions
(creating new internal quanti\[CapitalThorn]ers in the process), we can get
an
equivalent
prenex formula all of whose external quanti\[CapitalThorn]ers precede its
internal
quanti\[CapitalThorn]ers. But the same argument that shows that transfer
implies that
any
internal formula A is weakly equivalent to A_st also seems to
show that
given any internal formula in prenex form, standardizing an
intitial
segment
of the prenex gives rise to a weakly equivalent formula.
Indeed, to show
this last part, just use induction on the length of the
prenex segment and
the fact that the transfer principle implies the statement
that you get if
you replace any subset of the external generalizations that
occur in the
prenex of an instance of the transfer axiom with external
instantiations. I
have my doubts, though, about my being correct, for if I am
so, Why did
Nelson bother doing things the way he did, being careful to
push the
external generalizations to the left of the external
instantions (using the
logic trick that you can do this by de\[CapitalThorn]ning functions (which
inelegantly
require domains of de\[CapitalThorn]nition)), unless it be necessary? Do
you see a
mistake somewhere in my reasoning?
I refer not to NelsonÕs original paper, but to the chapter \
of
his
un\[CapitalThorn]nished
book at http://www.math.princeton.edu/~nelson/books.html .
===
Subject: Re: Internal Set Theory Uniqueness Principle
>> NelsonÕs theorem and proof can be extended to the
following result.
>> Let P(u,v_1,....,v_k) be a formula in Internal Set Theory
such that
>> the only occurrences of the predicate \
ŌstandardÕ in the
formula are
>> in quanti\[CapitalThorn]ers of the type for all standard z in V or for
some
>> standard z in V for some standard set V (V can differ from
quanti\[CapitalThorn]er
>> to quanti\[CapitalThorn]er). If y_1, ..., y_k are standard, and there is
a unique
>> x such that P(x,y_1,...,y_k) holds, the unique value of x
is standard.
>> My question relates to what happens if for all standard z
in V and
>> for some standard z in V are relaxed to for all standard z
and
>> for some standard z. Is it still true that x must be
standard?
>IÕm thinking this is true because it seems to me every
external formula
can
>be reduced to a weakly equivalent internal formula, with no
proviso
needed.
>Indeed, just as in logic, we can arrange to put the external
formula into
an
>equivalent prenex form, whose prenex is a sequence of
internal and
>external quanti\[CapitalThorn]ers. Using the triviality and its dual that
for all x
>commutes with for all standard x and then also using
idealization to
push
>the external generalizations to the left of the internal
instantiations
and
>the external instantiations to the left of the internal
generaliztions
>(creating new internal quanti\[CapitalThorn]ers in the process), we can
get an
equivalent
>prenex formula all of whose external quanti\[CapitalThorn]ers precede its
internal
>quanti\[CapitalThorn]ers.
The dif\[CapitalThorn]culty with this is that it canÕt always \
be made to
work. If you
have A r E st s A st t Q(r,s,t,...) or E r A st s E st t
Q(r,s,t,...),
where ŌAÕ denotes for all, \
ŌEÕ denotes there exists, \
ŌstÕ
denotes
standard, and Q(r,s,t,...) is a formula in IST, then
idealization canÕt
be invoked since A st t Q(r,s,t...) and E st t Q(r,s,t,...)
are
external formulae. This makes it dif\[CapitalThorn]cult (impossible?) to
drag all the
quanti\[CapitalThorn]ers over standard quantities to the outside where
transfer can be
invoked.
I have a copy of a paper by V. Kanovei where he claims to
have a formula
in IST which canÕt be reduced to a formula in ZFC, \
speci\[CapitalThorn]cally
A F [(F is a function with domain Z and A st n in Z (F(n) is
standard))
=> E st G (G is a function with domain Z and A st n in Z
(F(n) = G(n))].
>But the same argument that shows that transfer implies that
any
>internal formula A is weakly equivalent to A_st also seems
to show that
>given any internal formula in prenex form, standardizing an
intitial
segment
>of the prenex gives rise to a weakly equivalent formula.
Indeed, to show
>this last part, just use induction on the length of the
prenex segment and
>the fact that the transfer principle implies the statement
that you get if
>you replace any subset of the external generalizations that
occur in the
>prenex of an instance of the transfer axiom with external
instantiations.
I
>have my doubts, though, about my being correct, for if I am
so, Why did
>Nelson bother doing things the way he did, being careful to
push the
>external generalizations to the left of the external
instantions (using
the
>logic trick that you can do this by de\[CapitalThorn]ning functions (which
inelegantly
>require domains of de\[CapitalThorn]nition)), unless it be necessary? Do
you see a
>mistake somewhere in my reasoning?
As I remarked above, we canÕt handle cases like A r E st s A
st t or
E r A st s E st t. In the case where we have A r E st s in V
A st t
in W
Q(r,s,t,...) or E r A st s in V E st t in W Q(r,s,t,...),
where V and
W are standard and Q(r,s,t,...) is an internal formula, it is
easy to
handle since the A st and the E st can swapped in the manner
that
you
mention above, and then idealization can be applied.
>I refer not to NelsonÕs original paper, but to the chapter
of his
un\[CapitalThorn]nished
>book at http://www.math.princeton.edu/~nelson/books.html .
David
And all dared to brave unknown terrors, to do mighty deeds,
to boldly split in\[CapitalThorn]nitives that no man had split before -
and thus was the Empire forged.
-----
===
Subject: Re: Internal Set Theory Uniqueness Principle
> >require domains of de\[CapitalThorn]nition)), unless it be necessary? Do
you see a
> >mistake somewhere in my reasoning?
> As I remarked above, we canÕt handle cases like A r E st s
A st t or
> E r A st s E st t. In the case where we have A r E st s in
V A st t
in
W
> Q(r,s,t,...) or E r A st s in V E st t in W Q(r,s,t,...),
where V
and
> W are standard and Q(r,s,t,...) is an internal formula, it
is easy to
> handle since the A st and the E st can swapped in the
manner that
you
> mention above, and then idealization can be applied.
> >I refer not to NelsonÕs original paper, but to the \
chapter
of his
un\[CapitalThorn]nished
> >book at http://www.math.princeton.edu/~nelson/books.html .
> David
> And all dared to brave unknown terrors, to do mighty deeds,
> to boldly split in\[CapitalThorn]nitives that no man had split before -
> and thus was the Empire forged.
> -----
OK, yes, now I see. I stupidly forgot that idealization
required an
internal
formula. I suppose NelsonÕs algorithm works because a
sequence of n
external
generalizatons is like an external generalization of a
sequence with n
terms, which when between an internal formula and an internal
instantiation
can jump over the internal instantiation because of
idealization.
===
Subject: Re: Internal Set Theory Uniqueness Principle
>> >require domains of de\[CapitalThorn]nition)), unless it be necessary?
Do you see a
>> >mistake somewhere in my reasoning?
>> As I remarked above, we canÕt handle cases like A r E st \
s
A st t or
>> E r A st s E st t. In the case where we have A r E st s in
V A st
t in
>> Q(r,s,t,...) or E r A st s in V E st t in W Q(r,s,t,...),
where V
and
>> W are standard and Q(r,s,t,...) is an internal formula, it
is easy to
>> handle since the A st and the E st can swapped in the
manner that
you
>> mention above, and then idealization can be applied.
>> >I refer not to NelsonÕs original paper, but to the
chapter of his
>un\[CapitalThorn]nished
>> >book at http://www.math.princeton.edu/~nelson/books.html .
>> David
>> And all dared to brave unknown terrors, to do mighty deeds,
>> to boldly split in\[CapitalThorn]nitives that no man had split before -
>> and thus was the Empire forged.
>> -----
>OK, yes, now I see. I stupidly forgot that idealization
required an
internal
>formula.
I made the same mistake earlier when I attempted a proof in
the general
case. I managed to get to the point where I had concluded
that the unique
value under the general conditions must be standard, and then
when I went
back, I realized that I had been applying idealization willy
nilly to
external formulae, and I had to throw out all of my work.
>I suppose NelsonÕs algorithm works because a sequence of n
external
>generalizatons is like an external generalization of a
sequence with n
>terms,
ThatÕs correct. The statement that a set is such a sequence,
and that
its components are speci\[CapitalThorn]ed sets is internal, and so is
allowable.
>which when between an internal formula and an internal
instantiation
>can jump over the internal instantiation because of
idealization.
David
Men were real men, women were real women, and
small furry creatures from Alpha Centauri were
real small furry creatures from Alpha Centauri.
-----
===
Subject: Re: 3 Questions on Z[sqrt(n)]
days. My association with the Department is that of an
alumnus.
>>>If Z[sqrt(n)] is _not_ the ring of algebraic integers in
Q(sqrt(n))
>>>then it is not integrally closed, and so cannot be a UFD.
>>>So for the purpose of your query, the only cases of
interest
>>>are when Z[sqrt(n)] _is_ the ring of algebraic integers,
>>>ie when n is square-free and n is not = 1 mod 4.
>> Well, there are cases where Z[sqrt(n)] ->is<- the ring of
>> integers of Q(sqrt(n)), but n is not square free.
>IÕm sure you are right, as you usually are!
Hmmm... Maybe not. I think my error was to confuse the fact
that
Q(sqrt(n)) would equal Q(sqrt(m)), and conclude incorrectly
that
Z[sqrt(n)] would equal Z[sqrt(m)], where we write n = d^2*m
with m
square free. I think you were right. Sorry about that...
(And just a few hours ago, in another newsgroup, I said that
1976+30 =
1996... )
--
ItÕs not denial. IÕm just very selective \
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Polynomial sequence
Does anyone recognize this sequence of polynomials
(over F_2).
1
x
x^2+x+1
x^3
x^4+x^3+1
x^5+x^3+x
x^6+x^5+x^4+x^3+x^2+x+1
x^7
x^8+x^7+1
etc.
Bart
===
Subject: Re: Polynomial sequence
> Does anyone recognize this sequence of polynomials
> (over F_2).
> x^2+x+1
> x^3
> x^4+x^3+1
> x^5+x^3+x
> x^6+x^5+x^4+x^3+x^2+x+1
> x^7
> x^8+x^7+1
> etc.
f_n = (x + 1) f_(n - 1) + 1?
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Polynomial sequence
>> Does anyone recognize this sequence of polynomials
>> (over F_2).
>> 1
>> x
>> x^2+x+1
>> x^3
>> x^4+x^3+1
>> x^5+x^3+x
>> x^6+x^5+x^4+x^3+x^2+x+1
>> x^7
>> x^8+x^7+1
>> etc.
>f_n = (x + 1) f_(n - 1) + 1?
aka
n
(x+1) - 1
f = ----------
n x
Using the Lucas Correspondence Theorem
http://mathworld.wolfram.com/LucasCorrespondenceTheorem.html
which is proven more generally for multinomials at
http://www.whim.org/nebula/math/lucascorrespondence.html
it is easy to compute these polynomials. Let Ō&Õ \
be the
bitwise ŌandÕ
operator. Then, for m > 0, the coef\[CapitalThorn]cient of x^{m-1} in f_n
is 1 iff
n & m = m. This is why f_{2^n} has only one term, x^{2^n-1},
while all
2^n-1 terms appear in f_{2^n-1}.
For example, knowing that 100 = 1100100 base 2, we get that
99 95 67 63 35 31 3
f = x + x + x + x + x + x + x
100
since 1 more than each exponent has the binary expansion
100 = 1100100
96 = 1100000
68 = 1000100
64 = 1000000
36 = 100100
32 = 100000
4 = 100
That is, a 1 appears in the bits of exponent+1 only where a 1
appears in
the bits of n.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
===
Subject: Re: Polynomial sequence
Originator: anno4000@lublin.zrz.tu-berlin.de (Anno Siegel)
> >> Does anyone recognize this sequence of polynomials
> >> (over F_2).
> >>
> >> 1
> >>
> >> x
> >>
> >> x^2+x+1
> >>
> >> x^3
> >>
> >> x^4+x^3+1
> >>
> >> x^5+x^3+x
> >>
> >> x^6+x^5+x^4+x^3+x^2+x+1
> >>
> >> x^7
> >>
> >> x^8+x^7+1
> >>
> >> etc.
> >f_n = (x + 1) f_(n - 1) + 1?
> aka
> n
> (x+1) - 1
> f = ----------
> n x
> Using the Lucas Correspondence Theorem
> http://mathworld.wolfram.com/LucasCorrespondenceTheorem.html
> which is proven more generally for multinomials at
> http://www.whim.org/nebula/math/lucascorrespondence.html
> it is easy to compute these polynomials. Let \
Ō&Õ be the
bitwise ŌandÕ
> operator. Then, for m > 0, the coef\[CapitalThorn]cient of x^{m-1} in f_n
is 1 iff
> n & m = m. This is why f_{2^n} has only one term,
x^{2^n-1}, while all
> 2^n-1 terms appear in f_{2^n-1}.
> For example, knowing that 100 = 1100100 base 2, we get that
> 99 95 67 63 35 31 3
> f = x + x + x + x + x + x + x
> 100
> since 1 more than each exponent has the binary expansion
> 100 = 1100100
> 96 = 1100000
> 68 = 1000100
> 64 = 1000000
> 36 = 100100
> 32 = 100000
> 4 = 100
> That is, a 1 appears in the bits of exponent+1 only where a
1 appears in
> the bits of n.
This sequence can be generated by starting with k = m = 100,
then setting
k = (k - 1) & m until it vanishes.
Anno
===
Subject: Re: Polynomial sequence
>> Does anyone recognize this sequence of polynomials
>> (over F_2).
>> 1
>> x
>> x^2+x+1
>> x^3
>> x^4+x^3+1
>> x^5+x^3+x
>> x^6+x^5+x^4+x^3+x^2+x+1
>> x^7
>> x^8+x^7+1
>> etc.
>f_n = (x + 1) f_(n - 1) + 1?
... and thus with f_0 = 1, f_n = ((x+1)^(n+1)-1)/x.
Note also that (n choose m) = 1 mod 2 iff each binary digit
of n
is 1 whenever the corresponding binary digit of m is 1. Thus
e.g since 10 = 1010_2, (10 choose m) mod 2 = 1 only for m =
0, 2, 8,
and 10. Corresponding to this, f_9 = x^9 + x^7 + x^1.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Polynomial sequence
>>>Does anyone recognize this sequence of polynomials
>>>(over F_2).
>>>1
>>>x
>>>x^2+x+1
>>>x^3
>>>x^4+x^3+1
>>>x^5+x^3+x
>>>x^6+x^5+x^4+x^3+x^2+x+1
>>>x^7
>>>x^8+x^7+1
>>>etc.
>>f_n = (x + 1) f_(n - 1) + 1?
> ... and thus with f_0 = 1, f_n = ((x+1)^(n+1)-1)/x.
> Note also that (n choose m) = 1 mod 2 iff each binary digit
of n
> is 1 whenever the corresponding binary digit of m is 1. Thus
> e.g since 10 = 1010_2, (10 choose m) mod 2 = 1 only for m =
0, 2, 8,
> and 10. Corresponding to this, f_9 = x^9 + x^7 + x^1.
I spotted the same relation last night by sketching out a
grid of the
coef\[CapitalThorn]cients and noticing it represented a slightly clipped
sierpinksi
gasket.
Then if you know about pascals triangle mod 2 giving you the
gasket you can
see
the link.
Can I ask the original poster, from where did they pull these
polynomials
originally?
alex
===
Subject: Re: Polynomial sequence
>> ... and thus with f_0 = 1, f_n = ((x+1)^(n+1)-1)/x.
>> Note also that (n choose m) = 1 mod 2 iff each binary
digit of n
>> is 1 whenever the corresponding binary digit of m is 1.
Thus
>> e.g since 10 = 1010_2, (10 choose m) mod 2 = 1 only for m
= 0, 2, 8,
>> and 10. Corresponding to this, f_9 = x^9 + x^7 + x^1.
> I spotted the same relation last night by sketching out a
grid of the
> coef\[CapitalThorn]cients and noticing it represented a slightly clipped
sierpinksi
> gasket. Then if you know about pascals triangle mod 2
giving you the
> gasket you can see the link.
> Can I ask the original poster, from where did they pull
these
> polynomials originally?
I was noticing that if 2^k <= n < 2^{k+1}, then
f_n = x^{2^k}f_m + x^{2^k-1} + f_m
Where m=n-2^k.
I was treating a binary number as sequence b_n and looking at
its Z-transform X(b)(x). By the shift x->x+1, I got the
(shorter)
polynomials f_n and noticed the above. (Just looking at the
numerators of X(b).)
I was mostly wondering if this was a named or a studied
sequence. E.g., are they orthogonal in any sense?
Bart
===
Subject: The Ōchain of shadowsÕ
Please look at the attached pdf
http://www.geocities.com/complementarytheory/Roots-Chain.pdf
.
By this model we can see that 1 is the shadow of 2 and
2 is the shadow of 3.
I think that we can conclude that 3 is the shadow of 4
.. and so on.
In short, I am talking about roots which each one of them is
the
diagonal of its dimension level,
where each
n_dim
diagonal is the shadow of
[i:f3563246cc][b:f3563246cc]n[/b:f3563246cc][/i:f3563246cc]+1
_dim
diagonal.
We have a chain of shadows between in\[CapitalThorn]nitely many diagonals \
in
|[b:f3563246cc]N[/b:f3563246cc]| dimension levels.
Do you think that this Chain of Shadows has any
mathematical/physical meaning?
-------------------------------------------------------------
---------------
------------
Let us say that The chain of shadows is a
conjuncture.
Is there a possibility that there is no such a thing like a
diagonal,
in symmetrical elements (like a square in 2-d and a cube in
3-d),
which have more then 3 dimensions?
For example in 1-dim any line-segment of length 1 is its own
diagonal,
but in 2-dim and 3-dim the diagonal is the result of the of
its dimension level, and we get an irrational number, which is
different then the 1-dim case.
In 4-dim we again get a natural number as the result of the
diagonal
length.
In short, since in all cases we are talking about a diagonal
(which is
the root of its dimension level), then to disprove my
conjuncture we
have:
[b:f3563246cc]1)[/b:f3563246cc] To show that there is a clear
way to
conclude that diagonals which belong to dimension levels that
are
> 3, are not necessarily the root of their dimension.
[b:f3563246cc]2)[/b:f3563246cc] To show that there are
dimensions >
3 without diagonals.
If you have more ideas of how to disprove my conjuncture,
then Ill
be
glad to know them.
Here is again my diagram:
http://www.geocities.com/complementarytheory/Roots-Chain.pdf
-------------------------------------------------------------
---------------
------------------
Maybe these facts are aslo related to this conjuncture.
[u:f3563246cc][b:f3563246cc]Square
Numbers[/b:f3563246cc][/u:f3563246cc]
([url]http://www.krysstal.com/numbers.html[/url])
Square Numbers are integers that are the square of smaller
integers.
For example 4 is 22 and 9 is 32 so the \[CapitalThorn]rst few square
numbers are:
1 4 9 16 25 36 49 64 81 100 121 144 169
Note that the sequence of square numbers alternates between
odd and
even.
Another interesting fact is that this series of square
numbers can be
produced by adding successive odd numbers. For example, the
sum of
the \[CapitalThorn]rst two odd numbers (1, 3) is 1 + 3 = 4 (a square
number). The
sum of the \[CapitalThorn]rst three odd numbers (1, 3, 5) is 9 (another
sqauare).
This is shown in the table below:
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
1 + 3 + 5 + 7 + 9 + 11 = 36
1 + 3 + 5 + 7 + 9 + 11 + 13 = 49
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64
..
===
Subject: Re: The Ōchain of shadowsÕ
complementarytheory@yahoo-dot-com.no-spam.invalid (Doron
Shadmi)
> Please look at the attached pdf
> http://www.geocities.com/complementarytheory/Roots-Chain.pdf
> ....
> In short, I am talking about roots which each one of them
is the
> diagonal of its dimension level....
YouÕre talking about the diagonal of an n-dimensional unit
hypercube or measure polytope. Its vertices can be given
coordinates
(1,0,0,...,0), (0,1,0,...,)), (0,0,1,...,)), ......
(0,0,0,...,1),
and the n-dimensional analogue of the so-called Theorem of
Pythagoras
gives the length of its diagonal as
sqrt(1^2 + 1^2 + ... + 1^2) = sqrt(n) as you thought.
It may be easier to follow what IÕve just written if you \
\[CapitalThorn]rst
look at the special cases n = 2 and n = 3.
> ....
> Another interesting fact is that this series of square
numbers can be
> produced by adding successive odd numbers. For example, the
sum of
> the \[CapitalThorn]rst two odd numbers (1, 3) is 1 + 3 = 4 (a square
number). The
> sum of the \[CapitalThorn]rst three odd numbers (1, 3, 5) is 9 (another
square)....
Yes indeed. That fact also interested some Greek
mathematicians
of the 5th century B.C., who represented it by a pattern of
pebbles like
this:
* | * | * | * |
__| | | |
* * | * | * |
______| | |
* * * | * |
__________| |
* * * * | etc.
______________|
Can you see how each odd number borders a square to give a
larger
square? Those early Greek mathematicians used such pebble
patterns to
picture several properties of natural numbers.
Ken Pledger.
===
Subject: Re: The Ōchain of shadowsÕ
> .... Its vertices can be given coordinates
> (1,0,0,...,0), (0,1,0,...,)), (0,0,1,...,)), ......
(0,0,0,...,1),
> ....
Too much of the shift key. :-(
Those coordinates should have been
(1,0,0,...,0), (0,1,0,...,0), (0,0,1,...,0), ......
(0,0,0,...,1).
Ken Pledger.
===
Subject: Re: continuity of norm function in R^n
>> [...]
>> >(although my \[CapitalThorn]eld is nuclear medicine, really).
>> keen. iÕve read that if i take a few milligrams of
plutonium my
>> pain will go away in at most a few days. been wondering
where
>> to go for a second opinion... [plutonium isnÕt addictive,
is it?]
>As an evidence based medicine af\[CapitalThorn]cionado I have to tell you
that I
>donÕt know about any randomized controlled trial of
plutonium for this
>indication. A plutonium bomb might work, though: most people
stop
>complaining after a bomb is dropped on their head.
huh. iÕm tempted to ask if you could get me one of those, \
but
Ashcroft might not think it was funny...
************************
David C. Ullrich
sorry about the inelegant formatting - typing
one-handed for a few weeks...
===
Subject: Re: Arctan
>> Hi all,
>> Is the arctan function the same as appears on my scienti\[CapitalThorn]c
calc. as
>> tan^-1 or is that inverse tan? Or is inverse tan the same
as arctan?
>> Sorry if thatÕs unclear. IOW, does a key marked tan^-1
mean arctan?
>Those all mean the same thing.
>The -1 superscript in mathematics often means inverse,
>and arctan is another name for the same thing.
Sheesh. I did all this at school, but itÕs a
*looonnnnnnggggg* time
ago. :-(
--
What is now proved was once only imaginÕd. - William Blake,
1793.
===
Subject: Re: Arctan
Yes, it is. Or, inverse cosine.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Prime counting algorithms, speculation
<<snip>>
> holy cow, youÕre back to claiming itÕs \
-fast- compared to
existing
> methods? you really donÕt recall anything that happened \
the
last
> time you said this, right?
We had fun last time.
http://home.earthlink.net/~ewill3/math/primecounters/
index.html